Problem: The grades on a geometry midterm at Springer are normally distributed with $\mu = 74$ and $\sigma = 2.5$. Kevin earned a $73$ on the exam. Find the z-score for Kevin's exam grade. Round to two decimal places.
A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Kevin's exam grade by subtracting the mean $(\mu)$ from his grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{73 - {74}}{{2.5}}} $ ${ z \approx -0.40}$ The z-score is $-0.40$. In other words, Kevin's score was $0.40$ standard deviations below the mean.